Problem of Apollonius

First, we look for the first pair of circles that lie on one side of the given secant line. When dilating the given circle down to its center, the solution circles expand by the radius of the given circle. Therefore, the given lines are shifted by the radius of the circle.
This transforms the problem into solving for two lines and a point. We solve it, for example, using circular inversion (for a detailed explanation and other solution methods, see: https://apollonea.cz/en/pll/two-diverging-lines-with-a-point-between-them/).
The centers of the found circles are already the centers of the solutions. We reduce the radii of the found circles by the radius of the given circle, thus obtaining the first two solutions.
Next, we search for circles on the opposite side of the given secant line. When dilating the given circle down to its center, the solution circles again expand by the radius of the given circle, and the given lines are shifted by this radius as well.
Again, the problem is transformed into solving for two lines and a point. We solve it using circular inversion.
The centers of the found circles are the centers of the solutions. We reduce the radii of the found circles by the radius of the given circle, thus obtaining the second pair of solutions.
The problem has four solutions.

The centers of all circles tangent to both given lines lie on the angle bisectors between the lines.
The centers of circles tangent to the line and the circle lie on a pair of parabolas, where the focus is the center of the circle, and the directrix lines are parallels to the given line. These parallels are at a distance from the given line equal to the radius of the given circle.
The centers of the desired solution circles are located at the intersections of the parabolas and the angle bisectors.
The problem has four solutions.

Create the angle bisector. On them, the centres of the resulting circles lie.
Create tangent circles that are parallel to the lines from the task. One of the tangents will not be needed.
Create two lines that pass through the intersection of the lines from the task and the intersection of the tangents.
At the points where these lines intersect the given circle, the centers of the homothety are located.
Create the lines that pass from the center of the given circle to the centers of homothety. At the intersections of these lines with the angle bisectors lie the centers of the resulting circles. Note that although the line intersects both axes, the centers of the circles are only in the direction from the center to the point of homothety.
Draw circles from the centers found in the previous step, whose radius will be the distance from the center of the circles to the centers of the homothetal points.

Diverging Lines, One Is Intersecting A Circle